Question

A particle moves in a straight line and its position $x$ at time $t$ is given by $x^2=2+t$. Its acceleration is given by

• A. $\frac{-2}{x^3}$
• B. $-\frac{1}{4x^3}$
• C. $-\frac{1}{4x^2}$
• D. $\frac{1}{x^2}$

Answer 1

Answer: (B)

$x^2=t+2\Rightarrow \frac{1}{x^2}=\frac{1}{t+2}$ ,..(i)
$\Rightarrow x=\sqrt{t+2}$
$\Rightarrow \frac{dx}{dt}=\frac{1}{2}(t+2{)}^{\frac{1}{2}-1}$
$\Rightarrow \frac{dx}{dt}=\frac{1}{2}(t+2{)}^{-\frac{1}{2}}$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=\frac{1}{2}\left(-\frac{1}{2}\right)(t+2{)}^{\frac{1}{2}-1}$
$\Rightarrow a=-\frac{1}{4}(t+2{)}^{-\frac{3}{2}}$
$=-\frac{1}{4(t+2)}\times \frac{1}{(t+2{)}^{\frac{1}{2}}}$
$=-\frac{1}{4}\times \frac{1}{x^2}\times \frac{1}{x}$
$\Rightarrow a=-\frac{1}{4x^3}$